Prove That Continuous Function of Seqeunce Converges to F of Limit
This all comes down to the axiomatic and self-evident fact that when $\mathcal{A}$ and $\mathcal{B}$ are events, then $\mathbb{P}(\mathcal{A} \cup \mathcal{B}) \le \mathbb{P}(\mathcal{A}) + \mathbb{P}(\mathcal{B})$. Everything else is just manipulation with sets (which can be visualized with Venn diagrams).
Applying this requires decoding the notation, which in turn means remembering that most of it is shorthand for subsets of some sample space $\Omega$. At the outset, it is convenient to view the ordered pair $\mathrm{Z}_n = (\mathrm{X}_n, \mathrm{X})$ as a single random variable with values in the metric space $M = \mathbb{R}^k\times \mathbb{R}^k$.
I am going to reduce most of this notation to statements about the inverse images of functions from $\Omega$ to $\mathbb{R}$.
First, define
$$g^\prime:M \to \mathbb{R},\ g^\prime((\mathrm{X}, \mathrm{Y})) = g(\mathrm{X}) - g(\mathrm{Y}).$$
The functional composition
$$g^\prime \circ \mathrm{Z}_n:\Omega \to \mathbb{R},\ (g^\prime \circ \mathrm{Z}_n)(\omega) = g^\prime\left(Z_n\left(\omega\right)\right)$$
is a real-valued function on $\Omega$.
Let
$$\mathcal{A}_n = \{|g(\mathrm{X}_n) - g(\mathrm{X})| \gt \varepsilon\} = \{\omega\in\Omega\,:\, |g(\mathrm{X}_n(\omega)) - g(\mathrm{X}(\omega))| \gt \varepsilon\}.$$
This is the inverse image of the complement of the real interval $B(\varepsilon) = [-\varepsilon, \varepsilon]$:
$$\mathcal{A}_n = (g^\prime \circ \mathrm{Z}_n)^{-1}(\mathbb{R} \setminus B(\varepsilon)).$$
It's just a bunch of points in $\Omega$. (In words: $\mathcal{A}_n$ consists of all outcomes where the values of $g$ on $\mathrm{X}$ and $\mathrm{X}_n$ differ too much.)
(To simplify the notation, let's write set complements in $\mathbb{R}$ with overbars, as in
$$\bar B(\varepsilon) = \mathbb{R} \setminus B(\varepsilon)$$
for instance.)
Similarly, let
$$\mathcal{B} = \{\mathrm{X} \le K\} = \{\omega\in\Omega\,:\, \mathrm{X}(\omega) \le K\} = \mathrm{X}^{-1}(B(K))$$
(these are the outcomes where $\mathrm{X}$ is bounded by $K$) and
$$\mathcal{B}_n = \{\mathrm{X}_n \le K\} = \{\omega\in\Omega\,:\, \mathrm{X}_n(\omega) \le K\} = \mathrm{X}_n^{-1}(B(K))$$
(the outcomes where $\mathrm{X}_n$ is bounded by $K$).
Finally, writing
$$d:M\to\mathbb{R},\ d((\mathrm{X}, \mathrm{Y})) = |\mathrm{X}-\mathrm{Y}|,$$
set
$$\mathcal{C}_n = \{|\mathrm{X}_n-\mathrm{X}| \gt \gamma(\varepsilon)\} = \{\omega\in\Omega\,:\,|\mathrm{X}_n(\omega)-\mathrm{X}(\omega)| \gt \gamma(\varepsilon)\} = (d \circ \mathrm{Z}_n)^{-1}(\bar B(\gamma(\varepsilon)))$$
(the outcomes where $\mathrm{X}_n$ differs too much from $\mathrm{X}$).
All four sets are thereby seen to be inverse images of real-valued functions.
Bear in mind the relationship between inverse images of functions and complements. When $f:A\to B$ is any function between sets and $C\subset B$, then
$$ A \setminus f^{-1}(C) \subset f^{-1}(B\setminus C).$$
The proof is simple: on the left hand side are all elements $x\in A$ that do not get sent to $C$ by $f$. Since $f$--being a function--is defined on all elements of $A$, any such $x$ must get sent to the complement of $C$ in $B$, QED.
With those mechanical, set-theoretic preliminaries out of the way we can now interpret the text. In the statement after "since $g$ is uniformly continuous ..." we see
$$\{|g(\mathrm{X}_n - g(\mathrm{X})| \gt \varepsilon,\ \mathrm{X}\le K,\ \mathrm{X}_n \le K\} = \mathcal{A}_n \cap \mathcal{B} \cap \mathcal{B}_n$$
and on its right hand side is $\mathcal{C}_n$ itself. In the first step following "Hence" we see such sets as
$$``\mathrm{X} \gt K" = \mathrm{X}^{-1} (\bar B(K)) \supset \bar{\mathcal{B}}$$
and
$$``\mathrm{X}_n \gt K" = \mathrm{X}_n^{-1} (\bar B(K)) \supset \bar{\mathcal{B}_n}.$$
Their probabilities are summed. This suggests we consider the relationship between
$$\mathcal{A}_n \cap \mathcal{B} \cap \mathcal{B}_n \subseteq \mathcal{C}_n$$
and
$$\mathcal{A}_n \subseteq \mathcal{C}_n \cup \bar{\mathcal{B}} \cup \bar{\mathcal{B}_n}.$$
The clarity of this notation now makes it obvious that the first inclusion implies the second, because any $\omega\in\mathcal{A}_n$ that is not in $\mathcal{B}\cap\mathcal{B}_n$ will certainly lie in
$$\Omega \setminus (\mathcal{B}\cap\mathcal{B}_n) = \bar{\mathcal{B}} \cup \bar{\mathcal{B}_n}.$$
The left of the figure shows a Venn diagram with $\mathcal{A}_n \cap \mathcal{B} \cap \mathcal{B}_n$ in gray while the right of the figure shows $\mathcal{C}_n \cup \bar{\mathcal{B}} \cup \bar{\mathcal{B}_n} = \mathcal{C}_n \cup \left(\Omega \setminus \left(\mathcal{B} \cap \mathcal{B}_n\right)\right)$ in gray. The left gray region clearly is contained in the right gray region.
When we apply probabilities, the first inequality following "Hence" follows immediately because the probability of a union cannot exceed the sum of the probabilities, as noted at the outset.
The next inequality is derived in the same fashion: I leave the drawing of its Venn diagram to the interested reader.
Source: https://stats.stackexchange.com/questions/153885/a-continuous-function-of-a-sequence-of-random-vectors-converges-in-probability-t